Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 11

Multiply the numerator and denominator by the complex conjugate $1+i$: $(1+i)/(1-i)=(1+i{)}^2/2=i$, whose square is $-1$. Thus

The number $i{e}^{i\varphi }$ has modulus one and argument $\varphi +\pi /2$ modulo $2\pi$. Therefore its complex logarithm has valuesEvery value has real part zero. For the principal complex logarithm, choose the unique integer giving imaginary part in $(-\pi ,\pi ]$. The unwrapped value $\varphi +\pi /2$ is the principal imaginary part only when it already lies in that interval.

In polar form, a cube root of one has modulus one and argument $2\pi k/3$. The three roots of unity areThey are equally spaced on the unit circle.

Write $-8=8e^{i(\pi +2\pi k)}$. Its cube roots have modulus two and arguments $(\pi +2\pi k)/3$, for $k=0,1,2$. ThusEach cube equals $-8$, and a cubic polynomial has no further distinct roots.

For $b\ne 0$, write $z=bw$. The equation becomes $w^2+w+1=0$, whose roots are $(-1\pm i\sqrt{3})/2$, both of modulus one. Hence $|z|=|b|$. If $b=0$, the double root is $z=0$, giving the same relation. Therefore